Answer: $f(x) = x$, $f(x) = 0$.

Let us denote the given equation by (1):
$$f(x + y f(x + y)) = f(y^2) + x f(y) + f(x). \quad (1)$$

Substitute $x = y = 0$ into (1):
$$f(0) = 2f(0) \implies f(0) = 0.$$

Substitute $x = 0$ into (1):
$$f(y f(y)) = f(y^2). \quad (2)$$

Substitute $y = -x$ into (1):
$$f(x + (-x) f(x - x)) = f(x^2) + x f(-x) + f(x) \implies f(x) = f(x^2) + x f(-x) + f(x)$$
So,
$$f(x^2) = -x f(-x). \quad (3)$$

Now, substitute $x \to -x$ in (3):
$$f(x^2) = -(-x) f(-(-x)) = x f(x)$$
So,
$$f(x^2) = x f(x)$$
Comparing with (3):
$$x f(x) = -x f(-x) \implies f(-x) = -f(x). \quad (4)$$
So $f$ is odd.

Now, substitute $y = -y$ in (1):
$$f(x + (-y) f(x - y)) = f(y^2) + x f(-y) + f(x)$$
But $f(-y) = -f(y)$, so
$$f(x + (-y) f(x - y)) = f(y^2) - x f(y) + f(x)$$
Let us write this as:
$$f(x + y f(y - x)) = f(y^2) - x f(y) + f(x). \quad (5)$$

Now, consider the difference between (1) and (5):
$$f(x + y f(x + y)) - f(x + y f(y - x)) = 2x f(y). \quad (6)$$

Suppose that for some $x_0 \neq 0$, $f(x_0) = 0$. Substitute $y = x_0 - x$ into (1):
$$\begin{align*} f(x + (x_0 - x) f(x + x_0 - x)) &= f((x_0 - x)^2) + x f(x_0 - x) + f(x) \\ f(x) &= f((x_0 - x)^2) + x f(x_0 - x) + f(x) \\ f((x_0 - x)^2) &= -x f(x_0 - x) \end{align*}$$
But from (3):
$$f((x_0 - x)^2) = -(x_0 - x) f(x - x_0) = (x_0 - x) f(x_0 - x)$$
So,
$$-x f(x_0 - x) = (x_0 - x) f(x_0 - x) \implies f(x_0 - x) = 0, \forall x \in \mathbb{R}.$$
Thus, $f$ is identically zero, $f(x) = 0$ for all $x$.

If $f$ is not identically zero, then such $x_0 \neq 0$ cannot exist. Thus, from the condition $f(x_0) = 0$ it follows that $x_0 = 0$.

Now, let us prove injectivity. Suppose $\exists x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.
In (6), substitute $x = \frac{x_1 - x_2}{2}$, $y = \frac{x_1 + x_2}{2}$:
$$\begin{align*} f\left(\frac{x_1 - x_2}{2} + \frac{x_1 + x_2}{2} f(x_1)\right) - f\left(\frac{x_1 - x_2}{2} + \frac{x_1 + x_2}{2} f(x_2)\right) &= (x_1 - x_2) f\left(\frac{x_1 + x_2}{2}\right) \\ (x_1 - x_2) f\left(\frac{x_1 + x_2}{2}\right) &= 0 \end{align*}$$
So, either $x_1 = x_2$ (contradicts assumption), or $f\left(\frac{x_1 + x_2}{2}\right) = 0$. But as above, the only zero is at $0$, so $x_1 + x_2 = 0 \implies x_1 = -x_2$.
But then
$$f(x_2) = f(-x_1) = -f(x_1) = -f(x_2) \implies f(x_2) = 0 \implies x_2 = 0 \implies x_1 = 0$$
So, $f$ is injective.

Now, from (2):
$$f(y f(y)) = f(y^2)$$
But $f$ is injective, so $y f(y) = y^2 \implies f(y) = y$ for all $y$.

Thus, the only solutions are $f(x) = 0$ and $f(x) = x$.

It is easy to check that both satisfy the original equation.